How to resolve the algorithm Munchausen numbers step by step in the C# programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Munchausen numbers step by step in the C# programming language
Table of Contents
Problem Statement
A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n. (Munchausen is also spelled: Münchhausen.) For instance: 3435 = 33 + 44 + 33 + 55
Find all Munchausen numbers between 1 and 5000.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Munchausen numbers step by step in the C# programming language
Source Code 1:
- This code snippet calculates the sum of powers of digits for numbers from 1 to 5000 and prints the ones that equal the original number. It uses multiple Linq functions and a lambda expression.
Source Code 2:
- This code snippet checks if a given number is a "Munchhausen number," which is a number whose sum of the powers of its digits equals the number itself. It uses a cache to store precalculated powers of digits and iterates through numbers from 1 to 500,000,000 to find Munchhausen numbers.
Source Code 3:
- This code snippet has three different versions of a program that generates and prints Munchhausen numbers up to a certain limit. It uses nested loops and various mathematical calculations to check if the sum of the powers of the digits of a number equals the number itself.
In-depth Explanation:
Munchhausen Numbers: A Munchhausen number is a positive integer whose sum of the digits raised to the power of themselves is equal to the number itself. For example, 343 is a Munchhausen number because 3^3 + 4^4 + 3^3 = 343.
Source Code 1:
- The code uses the
Enumerable.Range(1,5000)method to generate a sequence of integers from 1 to 5000. - It then uses the
Wheremethod to filter the sequence and keep only the numbers that satisfy the condition:n == n.ToString().Sum(x => Math.Pow(toInt(x), toInt(x))). This checks if the number is equal to the sum of the powers of its digits. - Finally, it iterates through the filtered sequence and prints the numbers that satisfy the condition.
Source Code 2:
- The code initializes a cache array to store the powers of digits from 1 to 9.
- It then iterates through numbers from 1 to 500,000,000 and checks if each number is a Munchhausen number using the
IsMunchhausenmethod. - The
IsMunchhausenmethod calculates the sum of the powers of the digits of the number and compares it to the original number. It returnstrueif the number is a Munchhausen number andfalseotherwise.
Source Code 3:
- Version 1: This version uses nested loops to generate Munchhausen numbers and prints them. It calculates the sum of the powers of the digits of each number and checks if it equals the original number.
- Version 2: This version is similar to Version 1 but uses a different approach to generating Munchhausen numbers. It uses a table-driven method to quickly check if a number is a Munchhausen number.
- Version 3: This version further optimizes the table-driven approach used in Version 2 to generate Munchhausen numbers more efficiently.
Source code in the csharp programming language
Func<char, int> toInt = c => c-'0';
foreach (var i in Enumerable.Range(1,5000)
.Where(n => n == n.ToString()
.Sum(x => Math.Pow(toInt(x), toInt(x)))))
Console.WriteLine(i);
using System;
namespace Munchhausen
{
class Program
{
static readonly long[] cache = new long[10];
static void Main()
{
// Allow for 0 ^ 0 to be 0
for (int i = 1; i < 10; i++)
{
cache[i] = (long)Math.Pow(i, i);
}
for (long i = 0L; i <= 500_000_000L; i++)
{
if (IsMunchhausen(i))
{
Console.WriteLine(i);
}
}
Console.ReadLine();
}
private static bool IsMunchhausen(long n)
{
long sum = 0, nn = n;
do
{
sum += cache[(int)(nn % 10)];
if (sum > n)
{
return false;
}
nn /= 10;
} while (nn > 0);
return sum == n;
}
}
}
using System;
static class Program
{
public static void Main()
{
long sum, ten1 = 0, ten2 = 10; byte [] num; int [] pow = new int[10];
int i, j, n, n1, n2, n3, n4, n5, n6, n7, n8, n9, s2, s3, s4, s5, s6, s7, s8;
for (i = 1; i <= 9; i++) { pow[i] = i; for (j = 2; j <= i; j++) pow[i] *= i; }
for (n = 1; n <= 11; n++) { for (n9 = 0; n9 <= n; n9++) { for (n8 = 0; n8 <= n - n9; n8++) {
for (n7 = 0; n7 <= n - (s8 = n9 + n8); n7++) { for (n6 = 0; n6 <= n - (s7 = s8 + n7); n6++) {
for (n5 = 0; n5 <= n - (s6 = s7 + n6); n5++) { for (n4 = 0; n4 <= n - (s5 = s6 + n5); n4++) {
for (n3 = 0; n3 <= n - (s4 = s5 + n4); n3++) { for (n2 = 0; n2 <= n - (s3 = s4 + n3); n2++) {
for (n1 = 0; n1 <= n - (s2 = s3 + n2); n1++) {
sum = n1 * pow[1] + n2 * pow[2] + n3 * pow[3] + n4 * pow[4] +
n5 * pow[5] + n6 * pow[6] + n7 * pow[7] + n8 * pow[8] + n9 * pow[9];
if (sum < ten1 || sum >= ten2) continue;
num = new byte[10]; foreach (char ch in sum.ToString()) num[Convert.ToByte(ch) - 48] += 1;
if (n - (s2 + n1) == num[0] && n1 == num[1] && n2 == num[2]
&& n3 == num[3] && n4 == num[4] && n5 == num[5] && n6 == num[6]
&& n7 == num[7] && n8 == num[8] && n9 == num[9]) Console.WriteLine(sum);
} } } } } } } } }
ten1 = ten2; ten2 *= 10;
}
}
}
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